Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{z + 6}{z + 4} \div \dfrac{-4z^2 - 8z + 96}{-2z^2 + 10z + 72} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{z + 6}{z + 4} \times \dfrac{-2z^2 + 10z + 72}{-4z^2 - 8z + 96} $ First factor out any common factors. $t = \dfrac{z + 6}{z + 4} \times \dfrac{-2(z^2 - 5z - 36)}{-4(z^2 + 2z - 24)} $ Then factor the quadratic expressions. $t = \dfrac {z + 6} {z + 4} \times \dfrac {-2(z + 4)(z - 9)} {-4(z + 6)(z - 4)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(z + 6) \times -2(z + 4)(z - 9) } {(z + 4) \times -4(z + 6)(z - 4) } $ $t = \dfrac {-2(z + 4)(z - 9)(z + 6)} {-4(z + 6)(z - 4)(z + 4)} $ Notice that $(z + 6)$ and $(z + 4)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-2(z + 4)(z - 9)\cancel{(z + 6)}} {-4\cancel{(z + 6)}(z - 4)(z + 4)} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $t = \dfrac {-2\cancel{(z + 4)}(z - 9)\cancel{(z + 6)}} {-4\cancel{(z + 6)}(z - 4)\cancel{(z + 4)}} $ We are dividing by $z + 4$ , so $z + 4 \neq 0$ Therefore, $z \neq -4$ $t = \dfrac {-2(z - 9)} {-4(z - 4)} $ $ t = \dfrac{z - 9}{2(z - 4)}; z \neq -6; z \neq -4 $